Prove that $\phi : \mathbb{Q}(x,y) \to \mathbb{Q}(x,y),$ $f(x,y) \mapsto f(x+y,xy)$ is one-to-one

Question

Problem: The map $\phi : \mathbb{Q}(x,y) \to \mathbb{Q}(x,y), f(x,y) \mapsto f(x+y,xy)$ is a homomorphism. Prove that $\phi$ is one-to-one.

My Attempt: By the definition of the rational polynomials, $f(x,y) = \sum_{i,j}a_{i,j}x^iy^j$ which $\phi$ maps to $f(x+y,xy) = \sum_{i,j}a_{i,j}(x+y)^i(xy)^j$. We can then set $\sum_{i,j}a_{i,j}(x+y)^i(xy)^j = 0$. The only way for this statement to be true is to have $a_{i,j}=0\forall i,j$. This means that $\ker\phi = \{0\}$ since there is only the trivial solution to $\sum_{i,j}a_{i,j}(x+y)^i(xy)^j = 0$. Since $\ker\phi =0$, we then know that $\phi$ is one-to-one.

Is my proof for this okay, or do I need to say more? Particularly, do I need to prove that $(x+y)^i(xy)^j$ are linearly independent for all values of $i$ and $j$ to conclude that $a_{i,j}$ are all $0$?

Answer 1

Problem: "The map ϕ:Q(x,y)→Q(x,y),f(x,y)↦f(x+y,xy) is a homomorphism. Prove that ϕ is one-to-one."

Alternative "proof": Let $k$ be the field of rational numbers. Let $s_1:=x+y, s_2:=xy$. There is an action of the symmetric group $G:=S_2$ on $A:=k[x,y]$ and the invariant ring $A^G \cong k[s_1,s_2]$. The elements $s_1,s_2$ are algebraically independent hence there is an isomorphism

$$\phi: k[x,y] \cong k[s_1,s_2] \subsetneq k[x,y]$$

defined by

$$\phi(f(x,y)):=f(s_1,s_2).$$

The map $\phi$ is an isomorphism and induce a 1-1 map

$$\phi^*:k(x,y) \rightarrow k(s_1,s_2) \subsetneq k(x,y).$$

The elementary symmetric polynomials $s_1,s_2$ are algebraically independent over $\mathbb{Z}$:

$$\mathbb{Z}[x,y]^{S_2} \cong \mathbb{Z}[s_1,s_2]$$

and $\mathbb{Z}[s_1,s_2]$ is isomorphic to a polynomial ring. This construction generalize to several variables. You find a proof of this in any book on algebra (Lang's "Algebra").

Answer 2

You can reduce to the case of polynomials, because if $f(x,y)=\frac{f_1(x,y)}{f_2(x,y)}, g(x,y)=\frac{g_1(x,y)}{g_2(x,y)}$ where $f_i,g_i$ are polynomials, then we can show that $$f(x+y,xy)=g(x+y,xy)$$ if and only if $$f_1(x+y,xy)g_2(x+y,xy)=f_2(x+y,xy)g_1(x+y,xy).$$

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Lemma: (If $f(x,y),g(x,y)\in\mathbb Q[x,y]$ and for every $r_1,r_2\in\mathbb Q$ has $f(r_1,r_2)=g(r_1,r_2),$ then $f(x,y)=g(x,y).$

See later for proof.

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Now, if $f(x+y,xy)=g(x+y,xy)$ then for every complex values $x_1,y_1$ you have $f(x_1+y_1,x_1y_1)=g(x_1+y_1,x_1y_1).$

Now, for $r_1,r_2\in\mathbb Q,$ let $x_1,y_1$ be the two roots if $x^2-r_1x+r_2=0.$ Then $f(r_1,r_2)=f(x_1+y_1,x_1y_1)=g(x_1+y_1,x_1y_1)=g(r_1,r_2).$

So by the Lemma, if $f(x+y,xy)=g(x+y,xy)$ then $f(x,y)=g(x,y).$

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The underlying argument is that $\mathbb C\times\mathbb C, (z_1,z_2)\mapsto (z_1+z_2,z_1z_2)$ is onto, so if $(f-g)(z_1z_2,z_1+z_2)=0$ is true for all $z_1,z_2\in\mathbb C$ then $(f-g)(z_1,z_2)=0$ for all $z_1,z_2\in\mathbb C.$

The equivalent map $\mathbb Q\times\mathbb Q\to\mathbb Q\times \mathbb Q.$ is not onto. But evaluation of rational polynomials can be done for any values in any commutative ring containing the rationals, such as $\mathbb C.$

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Proof for Lemma. Let $h(x,y)=f(x,y)-g(x,y)$ and write:

$$h(x,y)=\sum_{i=0}^{n} h_i(x)y^i$$ where each $h_i(x)\in\mathbb Q[x]$ and $h_n(x)\neq 0.$ (We can do so if $f(x,y)\neq g(x,y).$)

For only finitely many $r\in\mathbb Q,$ we have $h_n(r)=0,$ so we can find an $r_1\mathbb Q$ so that $h_n(r_1)\neq 0.$

The $p(y)=h(r_1,y)=\sum_{i=0}^{n}h_n(r_1)y^i\in\mathbb Q[y],$ and $p(y)\neq 0,$ so there must be a $r_2\in\mathbb Q$ so that $h(r_1,r_2)=p(r_2)\neq 0.$

But then $f(r_1,r_2)-g(r_1,r_2)=h(r_1,r_2)\neq 0,$ so $f(r_1,r_2)\neq g(r_1,r_2).$ Thus contradicting our assumption.